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insert (5.1), interchange the integrals and then evaluate the inner integral.
2
Use the exercises below to complete the proof.


Exercises:
1. Check that in Theorems 24 and 25 it su¬ces to prove the case k = 1
and d = 0; the general case then follows through suitable changes of
variables.

2. For a sector S = S(0, ±) of in¬nite radius, let f ∈ A(1) (S, E ). For
su¬ciently large x0 > 0 and t > 0, show

e(t’u)(x0 +iy) ’ e(t’u)(x0 ’iy)
lim f (u) du = f (t), (5.5)
2πi (t ’ u)
y’∞ 0

integrating along the positive real axis; observe that u = t is a re-
movable singularity of the integrand.




5.4 A Di¬erent Representation for Borel Operators
Here we give a di¬erent integral representation for Borel operators of order
k > 1/2, using Mittag-Le¬„er™s function as a kernel. To do so, we ¬rst
prove the following well-known result upon the behavior of Mittag-Le¬„er™s
function as the variable z ’ ∞; for a more complete asymptotic analysis
of Es (z), compare the exercises below.
84 5. Integral Operators

Lemma 6 (Asymptotic of Mittag-Leffler™s function)
Let 0 < s < 2. Then for S’ = S(π, (2 ’ s)π) we have

z Es (z) ’’ ’1/“(1 ’ s), z ’ ∞,
S’ (5.6)

while for S+ = S(0, sπ) we have

Es (z) = s’1 exp[z 1/s ] + Es (z),
˜ (5.7)

˜
with z Es (z) bounded in S+ .

Proof: In the integral representation (B.19) (p. 233) one can deform the
path of integration γ (owing to Cauchy™s integral theorem) such that the
two radial parts follow the two rays arg w = ±(π+µ)/2, for arbitrarily small
µ > 0. Doing so, we ¬nd that for z with |π ’ arg z| ¤ (1 ’ s/2)π ’ sµ the
denominator ws ’ z never vanishes, so that we may interchange integration
and limit, by which we obtain (5.6), owing to (B.10) (p. 228). On the other
hand, for z ∈ S+ we can use the Residue Theorem to show

Es (z) = s’1 exp[z 1/s ] + Es (z),
˜ |z| > 1,

˜
where Es (z) has exactly the same integral representation (B.19), but for
˜
|z| > 1. From this representation, it is easy to see that z Es (z) remains
bounded as z ’ ∞ in S+ . 2

Remark 6: Since exp[z 1/s ] is bounded in the sectors sπ/2 ¤ ± arg z ¤
3sπ/2, we observe that (5.7) remains valid in the sector S(0, 3sπ). This
3
will be used in the proof of the next theorem.
For s as above, consider any sectorial region G = G(d, ±) of opening
± > sπ, and let f ∈ H(G, E ) be continuous at the origin. With k = 1/s
and γk („ ) as in Section 5.2, let

’1 dz
g(u) = Es (u/z) f (z) . (5.8)
2πi z
γk („ )


According to Lemma 6, the integral is absolutely convergent for u with
|d ’ arg u| < (± ’ sπ)/2 and „ ≈ arg u, and convergence is locally uniform
there, so that g(u) is analytic. We now show

Theorem 26 For 0 < s < 2 and any sectorial region G = G(d, ±) of
opening ± > sπ, let f ∈ H(G, E ) be continuous at the origin and de¬ne g
by (5.8). Then we have

g = Bk f, k = 1/s.
5.5 General Integral Operators 85

Proof: The integral for the Borel operator can be written as
’1 dz
k
kf (z) e(u/z)
(Bk f )(u) = .
2πi z
γk („ )

Hence, according to (5.7), resp. Remark 6, we have
’1 dz
˜
g(u) ’ (Bk f )(u) = Es (u/z) .
2πi z
γk („ )

Since the integrand is bounded in the interior of the path of integration,
we can use Cauchy™s integral theorem to conclude g ’ Bk f = 0. 2
The above representation of Borel operators, aside from being interesting
in its own right, also will serve as a starting point for more general integral
operators in the next section. Concerning the restriction of s to the interval
0 < s < 2, so that k = 1/s > 1/2, one should note that this is forced by the
fact that Mittag-Le¬„er™s function behaves di¬erently for other values of s.
In later chapters we shall see other results indicating that the value of 1/2
is special in the sense that some aspects of the theory of summability of
power series change when the parameter k is larger, resp. smaller, than 1/2.

Exercises: Let 0 < s < 2 and S’ = S(π, (2 ’ s)π).
1. For N ≥ 0, show
N ’1
1’N
z ’n /“(1 ’ sn),
Es (z; N ) ’
Es (z) = z (5.9)
n=1

ew wsN ’1 (ws ’ z)’1 dw.
with 2πi Es (z; N ) = γ

2. For Es (z; N ) as above, show z Es (z; N ) ’ ’1/“(1’sN ), as z ’ ∞ in
S’ . Interpret this, together with (5.9), as an asymptotic for Es (1/z).




5.5 General Integral Operators
In this section, we are going to study some generalizations of Laplace resp.
Borel operators. In detail, we are going to de¬ne pairs of integral operators,
one being the inverse of the other on suitable function spaces, and each pair
being associated with one particular “sequence of moments.” To do so, we
introduce pairs of functions serving as kernels for these operators.

Kernel Functions
A pair of C -valued functions e(z), E(z) will be called kernel
functions if for some k > 1/2 the following holds:
86 5. Integral Operators

• The function e(z) is holomorphic in S+ = S(0, π/k), and
z ’1 e(z) is integrable at the origin, meaning that the in-
x
tegral 0 0 x’1 |e(xei„ )| dx exists, for arbitrary x0 > 0 and
2k|„ | < π. Moreover, for every µ > 0 there exist constants
c, K > 0 such that

|e(z)| ¤ c exp[’(|z|/K)k ], 2k| arg z| ¤ π ’ µ. (5.10)

• For positive real z = x, the values e(x) are positive real.
• The function E(z) is entire and and of exponential growth
at most k. Moreover, in S’ = S(π, π(2 ’ 1/k)), i.e., the
complement of S+ in C , the function z ’1 E(1/z) is inte-
grable at the origin in the above sense.
• The functions e(z), E(z) are linked by the following mo-
ment condition: De¬ne the moment function corresponding
to the kernel e(z) by

Re u ≥ 0.
xu’1 e(x) dx,
m(u) = (5.11)
0

Note that the integral converges absolutely and locally
uniformly for these u, so that m(u) is holomorphic for
Re u > 0 and continuous up to the imaginary axis, and
the values m(x) are positive real numbers for x ≥ 0. Then
the power series expansion of E(z) equals

zn
E(z) = . (5.12)
m(n)
0

The number k will sometimes be referred to as the order of the
pair of kernel functions e(z), E(z).

Observe that the kernel E(z) is uniquely determined by e(z) because of
(5.12). Obviously, (5.10) implies k m(n) ¤ c K n “(n/k), n ≥ 1. On the
other hand, by assumption we have that E(z) is of exponential growth at
˜
most k, meaning by de¬nition |E(z)| ¤ c exp[K|z|]k , for su¬ciently large
˜
˜˜
c, K > 0. From the proof of Theorem 69 (p. 233) we conclude that this
ˆˆ ˆˆ
implies existence of c, K > 0 so that m(n) ≥ c K n “(1 + n/k), n ≥ 0. Hence
the moments m(n) are of order “(1+n/k) as n tends to in¬nity in the sense
that [m(n)/“(1 + n/k)]±1/n is bounded. In particular, this shows that the
order of a pair of kernel functions is uniquely de¬ned, and that the (entire)
kernel E(z) is exactly of exponential growth k, or in other words, is of
order k and ¬nite type. As a ¬rst example of such kernel functions we take
e(z) = k z k exp[’z k ]; in this case m(u) = “(1 + u/k) and E(z) = E1/k (z).
Other examples of interest will follow in the exercises and the next section.
5.5 General Integral Operators 87

With the help of any such pair of kernel functions of order k > 1/2, we
now de¬ne a pair of integral operators as follows:
Let S = S(d, ±) be a sector of in¬nite radius, and let f ∈ A(k) (S, E ),
then for |d ’ „ | < ±/2, the integral
∞(„ )
du
(T f )(z) = e(u/z) f (u) (5.13)
u
0

converges absolutely and locally uniformly for z with (4.1) (p. 61), for
su¬ciently large c > 0. As for Laplace operators, a change of „ results in
holomorphic continuation of T f . Hence we conclude that T f is holomorphic
in a sectorial region G(d, ± + π/k).
If G = G(d, ±) is a sectorial region of opening larger than π/k, and
f ∈ H(G, E ) is continuous at the origin, then we de¬ne, with γk („ ) as in
the de¬nition of Borel operators on p. 80,
’1 dz
(T ’ f )(u) = E(u/z) f (z) . (5.14)
2πi z
γk („ )

We conclude as in the proof of Theorem 24 (p. 82) that T ’ f is holomorphic
and of exponential growth at most k in S(d, ± ’ π/k).
In case e(z) = k z k exp[’z k ], the two integral operators coincide with
Laplace resp. Borel operators. Even in general they have many properties
in common with those classical operators, as we now show:

1. For f (u) = u» , with Re » > 0, hence f (u) continuous at the origin,
we have T f (z) = m(») z » ; to see this, make a change of variable in
(5.13) and use (5.11).

2. For f (u) = 0 fn un being entire and of exponential growth at most
k, the function T f is holomorphic for |z| < ρ, with su¬ciently small

ρ > 0, and (T f )(z) = 0 fn m(n) z n , |z| < ρ; to see this, check that
termwise integration of the power series expansion of f is justi¬ed.
3. For w = 0 and z = 0 so that |z/w| is su¬ciently small, it follows from
the above that
∞(„ )
w du
= e(u/z) E(u/w) . (5.15)
w’z u
0

This formula extends to values w = 0 and z = 0 for which both
sides are de¬ned. In particular, this is so for arg w = arg z modulo
2π, since then we can choose „ so that |„ ’ arg z| < π/(2k) and
|π ’ „ + arg w| < π(2 ’ 1/k), implying absolute convergence of the
integral, according to the properties of kernel functions.
4. For a sectorial region G = G(d, ±) of opening more than π/k, and
f ∈ H(G, E ) continuous at the origin, the composition h = T —¦ T ’ f
88 5. Integral Operators

is de¬ned. Interchanging the order of integration and then evaluating
the inner integral, using (5.15), implies

’1 f (w)
h(z) = dw = f (z),
w’z
2πi γk („ )

since γk („ ) has negative orientation. Hence we conclude that T ’ is
an injective integral operator, and T is its inverse. Note, however,
that we do not yet know that either operator is bijective; this will be
shown in Theorem 30 (p. 92).

5. For Re » > 0 and f (z) = z » , we conclude from (5.14) by a change of
variable u/z = w, and using Cauchy™s theorem to deform the path of
integration:

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